OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Guide to related sequences:
***** Part 1: initial values are a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6
A295357: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3)
A295358: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3)
A295359: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2*b(n-3)
A295360: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*b(n-3)
A295361: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - 3*b(n-3)
A295362: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) - b(n-3)
***** Part 2: initial values as shown
A295363: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295364: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4
A295365: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
A295366: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
A295367: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
For all of these sequences, a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = a(1) + a(0) + b(2) + b(1) + b(0) = 20
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 32; u = Table[a[n], {n, 0, z}] (* A295357 *)
v = Table[b[n], {n, 0, 10}] (* complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 21 2017
STATUS
approved